Two electrolytic cells containing CuSO_4 and | vedclass

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(B) According to Faraday's second law of electrolysis,for cells connected in series,the mass of substance deposited is proportional to its equivalent

Vedclass · Accepted Answer

$3.4$

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(B) According to Faraday's second law of electrolysis,for cells connected in series,the mass of substance deposited is proportional to its equivalent weight: $\frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}}$.Equivalent weight $E = \frac{	ext{Atomic weight}}{	ext{Valency}}$.For $Cu$ $(Cu^{2+})$: $E_{Cu} = \frac{63.57}{2} = 31.785$.For $Ag$ $(Ag^+)$: $E_{Ag} = \frac{107.88}{1} = 107.88$.Given $m_{Cu} = 1 \, mg$,we have $\frac{1}{m_{Ag}} = \frac{31.785}{107.88}$.$m_{Ag} = \frac{107.88}{31.785} \approx 3.394 \, mg \approx 3.4 \, mg$.

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